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Unstable Equilibrium

  1. Jun 14, 2013 #1
    1. The problem statement, all variables and given/known data
    If a beam with square cross-section and very low density is placed in water, it will turn one pair of its long opposite faces horizontal. This orientation,however, becomes unstable as we increase its density. Find the critical density when
    this transition occurs. The density of water is ρ v = 1000 kg/m3 .


    2. Relevant equations
    Lacking any idea,do not know how to start.


    3. The attempt at a solution
    I did not understand the question.Could someone clarify it and tell me the relevant equations.
     
  2. jcsd
  3. Jun 14, 2013 #2

    mfb

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    If you do not know how to start, draw a sketch.

    You should know equations for swimming objects.
     
  4. Jun 14, 2013 #3
    I am sorry.This wasn't quite helpful.Could you elaborate?
     
  5. Jun 14, 2013 #4

    haruspex

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    You understand the physical arrangement in the first sentence, yes?
    If it becomes unstable, in what way will the orientation be likely to change?
    How would you go about determining whether that change will happen?
     
  6. Jun 14, 2013 #5

    D H

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    Hint, in the form of a couple of questions:
    • What distinguishes a stable equilibrium from an unstable one?
    • What happens when the system is very close to but not at the equilibrium point?
     
  7. Jun 15, 2013 #6
    Well,if the second differential of the potential is negative,then the system will be unstable.But,how to find the equilibrium point.also,how to find the potential energy.
     
  8. Jun 15, 2013 #7

    D H

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    Think a bit more primitively then: Forces and torques. You can either solve the problem directly with a force/torque approach, or you can use that to lead you toward the potential function.
     
  9. Jun 15, 2013 #8

    haruspex

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    The first step is this question I asked before: If it becomes unstable, in what way will the orientation be likely to change? Once you have that, you can suppose it changes by small amount in that direction and analyse the forces that will then apply. (The differential you referred to is wrt that change.)
     
  10. Jun 16, 2013 #9
    Okay,so then the initial orientation,as i understand,is at first that the beam is parallel to the surface of water.as its density is increased,slowly its centre of gravity will come down,once it is just below the surface of water,the orientation will become unstable.So the answer is 500kg/m^3.is it?could someone verify the answer.
     
  11. Jun 16, 2013 #10

    mfb

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    Is that a guess, or did you calculate it?
     
  12. Jun 16, 2013 #11

    D H

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    Correct.

    Incorrect.

    That is not the correct answer. Since you just stated a result rather than showing your work I don't know where you went wrong.
     
  13. Jun 16, 2013 #12
    Frankly speaking,it was a guess.Since I still cannot relate how the configuration will become unstable.Since it has square orientation,if the density continues to increase,it will just sink,won't it?
    Let me try to tell the forcess.First, of course, a downward force due to weight.
    Second,A downward pressure due to Atmosphere.
    Third,An upward pressure due to the displaced water,the bouyant force.
    Are there any other forces?
     
  14. Jun 16, 2013 #13

    D H

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    Yes, it will sink a bit. That's not what the question is asking. It's asking about the stable orientation of the beam. If the density is very low the beam will float with two faces parallel to the surface of the water. One of those two faces will be entirely submerged, the other entirely above the surface. At some point as density increases this configuration ceases to be stable. Some other configuration becomes the stable one. There are in fact four different stable orientations, the aforementioned flat top orientation plus three others. Which one is the stable one depends on density. What's perhaps most surprising as that as density increases even further, the flat top orientation once again becomes the stable one. A beam that is just slightly less dense than water will float in that flat top orientation.

    The question is asking about that first transition point, where the flat top orientation stops being stable. You don't even need to find the which orientation is stable. You just have find the transition point.
     
  15. Jun 16, 2013 #14
    What about the forces I mentioned?Are there any other forces?
     
  16. Jun 16, 2013 #15

    D H

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    You can safely ignore atmospheric pressure. It too is a buoyancy force, and it's very small compared to those other two for something with a density that is any sizable fraction of that of water.
     
  17. Jun 16, 2013 #16
    Okay,from what i Understand,due to changes in the centre of buoyancy there will be an unbalanced moment,as a result the stable config will change.Is my assumption correct?
     
  18. Jun 16, 2013 #17

    D H

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    Correct. So where are those forces applied? What distinguishes an unbalanced torque from a restoring torque?
     
  19. Jun 16, 2013 #18
    Well,the forces are applied at the centre of gravity and centre of buoyancy.If its stable,for small changes it will come back to its original state.But if its unstable,Then due to the unbalanced torque it will not come back.But how to determine the new centre of buoyancy,After i have tilted it a little?And how many axes should i consider when taking calculations?
     
  20. Jun 16, 2013 #19

    D H

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    Small angle approximation is always a good way to start. As far as which axes you need to be concerned with, there's only one, and it's the obvious one.
     
  21. Jun 16, 2013 #20
    Oh,yeah,the axes chosen as the centre of gravity as origin.Could u suggest me some books that would strengthen my concepts regarding the problem.
     
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